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erik [133]
3 years ago
10

Freddie drives his car 10 km North. He stops for gas and then drives 5 km South.

Physics
1 answer:
fgiga [73]3 years ago
4 0

Answer:

5 km

Explanation:

displacement is the straight line distance from a particular point. since north and south are in opposite directions, it means that he has taken a u-turn

thus, 10 - 5 = 5km displacement

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an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo
Otrada [13]

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

F = ma

9000 = 6000 a

a = 1.5 m/s^2

now for finding minimum stopping distance of the truck

v_f^2 - v_i^2 = 2a d

0^2 - 15^2 = 2(-1.5) d

d = 75 m

4 0
3 years ago
PLEASEEEE HELPPPPPPPPPPPP
lesantik [10]

Answer:

240 meter

Explanation:

d= s*d=60m/s * 4sec=240 meter

7 0
2 years ago
Diffraction supports the:<br><br> A. wave theory of light.<br><br> B. particle theory of light.
Artemon [7]
The answer is a hope its helps you
8 0
3 years ago
Read 2 more answers
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini
kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

  • Mass of cart one M1  = 150kg
  • Initial Velocity U1 = 8m/s
  • Final VelocityV1 = 5 m/s

Mass of cart two M2  = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0
2 years ago
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